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This question already has an answer here:

Based on the definition of $e: = \lim_{x\to\infty} \left(1+\frac1x \right)^x$, how can we show that

$$\lim_{x\to \infty}\left( 1-\frac{\lambda}{x} \right)^x = e^{-\lambda}?$$

So far I've tried changing variables, $\eta = \frac{-x}{\lambda}$, so $=\lim_{\eta \to -\infty}\left( \left( 1 + \frac1\eta \right)^\eta \right)^{-\lambda}$. But then we would need to show $\lim_{\eta \to -\infty}\left( 1 + \frac1\eta \right)^\eta =e$.

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marked as duplicate by Guy Fsone, Dando18, mechanodroid, J. M. is a poor mathematician, rlartiga Nov 9 '17 at 21:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Hint: $$\left(1+\frac1{\eta}\right)^\eta = \left(\frac1{\left(1+\frac1{-(\eta+1)}\right)^{-(\eta+1)}}\right)^{-\eta/(\eta+1)}$$ and $-(\eta+1) \to +\infty$ as $\eta \to -\infty$.

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  • $\begingroup$ Very cool! I wonder why I never did this in Advanced Calculus... $\endgroup$ – Eric Auld Oct 13 '13 at 20:02
  • $\begingroup$ Is this an example of a more general strategy of some kind? How did you figure out to do this? $\endgroup$ – Eric Auld Oct 13 '13 at 22:29
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$\newcommand{\abs}[1]{\left\vert #1\right\vert}$ When $\lambda = 0$, the result is trivially true: $1 = {\rm e}^{0}$. Let's consider the cases $\lambda \not= 0$:

  1. $\large\lambda < 0$ $$ \lim_{x\to \infty}\left(1 - {\lambda \over x}\right)^{x} = \lim_{x\to \infty}\left[% \left(1 + {\abs{\lambda} \over x}\right)^{x/\abs{\lambda}}\right]^{\abs{\lambda}} = \lim_{x\to \infty}\left[% \left(1 + {1 \over x}\right)^{x}\,\right]^{\abs{\lambda}} = {\rm e}^{\abs{\lambda}} = {\rm e}^{-\lambda} $$
  2. $\large\lambda > 0$ $$ \lim_{x\to \infty}\left(1 - {\lambda \over x}\right)^{x} = \lim_{x\to \infty}\left[% \left(1 - {1 \over x/\lambda}\right)^{-x/\lambda}\right]^{-\lambda} = \lim_{x\to \infty}\left[% \left(1 - {1 \over x}\right)^{-x}\right]^{-\lambda} = {\rm e}^{-\lambda} $$

Otherwise, $$ \lim_{x\to \infty}\left(1 - {\lambda \over x}\right)^{x} = \lim_{x\to \infty} \exp\left(\vphantom{\LARGE A^{A}}\,x\ln\left(1 - {\lambda \over x}\right)\right) = \lim_{x\to \infty} \exp\left(\vphantom{\LARGE A^{A}}\,x\left[-{\lambda \over x}\right]\right) = {\rm e}^{-\lambda} $$

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